---
title: "Resistance of alloy sheet along length and thickness"
subject: "Basic Electrical Engineering"
chapter: 1
key: "semester_1/basic_electrical_engineering:resistivity-alloy-sheet-resistance"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/resistivity-alloy-sheet-resistance
site: BCT Engineering
---

# Resistance of alloy sheet along length and thickness

Basic Electrical Engineering · Chapter 1: General Electric System

- Asked in: 2069 Chaitra, 6 marks
- Video: [The resistivity of a ferric-chromium-aluminium alloy is51 × 10−8 Ω-m. BASIC ELECTRICAL ENGINEERING](https://www.youtube.com/watch?v=4viFZGSVK4s) — Target Nasa
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/resistivity-alloy-sheet-resistance
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-1

## Question

The resistivity of a metal alloy is 50×10⁻⁸Ω-m. A sheet of material 15 cm long, 6 cm wide and 0.014 cm thick. Calculate the resistance in the direction: (a) along the length and (b) along the thickness.

## Short answer

R = ρ l/A. (a) Along length: l=0.15 m, A=8.4×10⁻⁶ m², R≈8.93 mΩ. (b) Along thickness: l=1.4×10⁻⁴ m, A=9×10⁻³ m², R≈7.78 nΩ.

## Answer

Use R = ρ l/A, where *l* is the length of the current path and *A* the cross-section at right angles to the current.

#### Data
ρ = 50 × 10⁻⁸ Ω·m, length = 15 cm = 0.15 m, width = 6 cm = 0.06 m, thickness = 0.014 cm = 1.4 × 10⁻⁴ m

#### (a) Along the length
Path length l = 0.15 m
Area A = width × thickness = 0.06 × 1.4 × 10⁻⁴ = 8.4 × 10⁻⁶ m²

R = 50 × 10⁻⁸ × 0.15/(8.4 × 10⁻⁶) = 8.93 × 10⁻³ Ω

#### (b) Along the thickness
Path length l = 1.4 × 10⁻⁴ m
Area A = length × width = 0.15 × 0.06 = 9 × 10⁻³ m²

R = 50 × 10⁻⁸ × 1.4 × 10⁻⁴/(9 × 10⁻³) = 7.78 × 10⁻⁹ Ω

**Answer: (a) ≈ 8.93 mΩ; (b) ≈ 7.78 nΩ**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In R = rho l / A, what do l and A represent?
   A. Longest side and total surface area
   B. Length of the current path and cross-section at right angles to the current
   C. Width and thickness always
   D. Length of the sheet and its volume
   Answer: B. l is along the current direction and A is perpendicular to it, so they change with direction. (Key point: Path length and area)

2. Along the length the area used is:
   A. length x width = 9e-3 m^2
   B. width x thickness = 8.4e-6 m^2
   C. length x thickness
   D. thickness only
   Answer: B. Current flows along the 15 cm, so A = 0.06 x 1.4e-4 = 8.4e-6 m^2, giving about 8.93 mOhm. (Key point: Along the length)

3. Along the thickness the path length and area are:
   A. 0.15 m and 8.4e-6 m^2
   B. 1.4e-4 m and 9e-3 m^2
   C. 1.4e-4 m and 8.4e-6 m^2
   D. 0.06 m and 9e-3 m^2
   Answer: B. Path is the thickness 1.4e-4 m, area is length x width = 0.15 x 0.06 = 9e-3 m^2, giving about 7.78 nOhm. (Key point: Along the thickness)

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